Hey,

I have been working on this question for the past couple days (searching the internet, searching for examples in textbook etc.) but I cannot figure out how to answer this question. I will likely have follow up questions to your solution, so please keep an eye on this thread following helping me!

Question

Determine a scalar equation of a plane that contains the line r = (3, -1, 1) + t(2, 4, -5) and whose z-intercept is -2.

Solution

The textbook gives a hint in the back of the book that I need 2 direction vectors in order to find the solution. The question is giving me one direction vector "(2, 4, -5)" but I am unsure of how to find the second one. After I find the second direction vector (which I am unsure how to do) I find the cross product of the two direction vectors to find the normal. After having the normal I think I need a point on the plane The z intercept is listed, so I know a point is (0, 0, -2), right?

Knowing this I can simply sub into the D = -Ax - By - Cz. (which A B C and are from the normal and the x y z are from the point on the plane)

This is how I am understanding what I need to do. PLEASE correct me if I am mistaken here! I am unsure how to do certain steps in my explanation, which is what I am needing help with.

Thank you for helping!!!

I have been working on this question for the past couple days (searching the internet, searching for examples in textbook etc.) but I cannot figure out how to answer this question. I will likely have follow up questions to your solution, so please keep an eye on this thread following helping me!

Question

Determine a scalar equation of a plane that contains the line r = (3, -1, 1) + t(2, 4, -5) and whose z-intercept is -2.

Solution

The textbook gives a hint in the back of the book that I need 2 direction vectors in order to find the solution. The question is giving me one direction vector "(2, 4, -5)" but I am unsure of how to find the second one. After I find the second direction vector (which I am unsure how to do) I find the cross product of the two direction vectors to find the normal. After having the normal I think I need a point on the plane The z intercept is listed, so I know a point is (0, 0, -2), right?

Knowing this I can simply sub into the D = -Ax - By - Cz. (which A B C and are from the normal and the x y z are from the point on the plane)

This is how I am understanding what I need to do. PLEASE correct me if I am mistaken here! I am unsure how to do certain steps in my explanation, which is what I am needing help with.

Thank you for helping!!!

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